package org.leetcode.middle.leetcode583;

public class Solution {


    public int minDistance(String word1, String word2) {
        int n1 = word1.length();
        int n2 = word2.length();

        int[][] dp = new int[n1 + 1][n2 + 1];

        for (int i = 0; i <= n1; i++) {
            dp[i][0] = i;
        }

        for (int j = 0; j <= n2; j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i <= n1; i++) {
            for (int j = 1; j <= n2; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    //dp[i - 1][j - 1] + 2 表示同时删除当前i和j指向的字符，需要额外操作两次
                    //dp[i - 1][j] + 1 表示删除当前i指向的字符，需要额外操作一次
                    //dp[i][j - 1] + 1 表示删除当前j指向的字符，需要额外操作一次
                    dp[i][j] = Math.min(dp[i - 1][j - 1] + 2, Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
                }
            }
        }

        return dp[n1][n2];
    }

    //求两个word的最长公共子序列长度，再计算每个word删除到最长公共子序列的次数
    public int minDistance2(String word1, String word2){

        int n1 =word1.length();
        int n2 =word2.length();

        int [][] dp = new int[n1+1][n2+1];

        int maxLength = 0;

        for (int i = 1; i <=n1; i++) {
            for (int j = 1; j <=n2 ; j++) {
                if (word1.charAt(i-1)==word2.charAt(j-1)){
                    dp[i][j]=dp[i-1][j-1]+1;
                }else {
                    dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
                }

                maxLength=Math.max(maxLength,dp[i][j]);
            }
        }


        return n1+n2-2*maxLength;
    }
}
